To determine the ratio of the kinetic energy of the $$n=2$$ electron for the H atom to that of the He$$^+$$ ion, we need to consider the Bohr model of the atom. According to the Bohr model, the kinetic energy (KE) of an electron in a hydrogen-like ion in the nth orbit can be given by:

$$ KE_n = \frac{1}{2} m v_n^2 = \frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2} $$

where:

• $ m $ is the mass of the electron
• $ Z $ is the atomic number of the nucleus
• $ e $ is the charge of the electron
• $ \epsilon_0 $ is the permittivity of free space
• $ h $ is Planck's constant
• $ n $ is the principle quantum number, which is 2 in this case

For a hydrogen atom (H) ($ Z = 1 $) in the $ n = 2 $ state:

$$ KE_{H (n=2)} = \frac{1^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{e^4 m}{32 \epsilon_0^2 h^2} $$

For a singly ionized helium ion (He$$^+$$) ($ Z = 2 $) in the $ n = 2 $ state:

$$ KE_{He^+ (n=2)} = \frac{2^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{4 e^4 m}{32 \epsilon_0^2 h^2} = \frac{e^4 m}{8 \epsilon_0^2 h^2} $$

To find the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$$^+$$ ion, we divide the kinetic energies:

$$ \text{Ratio} = \frac{KE_{H (n=2)}}{KE_{He^+ (n=2)}} = \frac{\frac{e^4 m}{32 \epsilon_0^2 h^2}}{\frac{e^4 m}{8 \epsilon_0^2 h^2}} = \frac{1}{4} $$

Therefore, the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$$^+$$ ion is:

Option A: $$\frac{1}{4}$$